# Check if one string can be converted to other using given operation

Given two strings S and T of same length. The task is to determine whether or not we can build a string A(initially empty) equal to string T by performing the below operations.

1. Delete the first character of S and add it at the front of A.
2. Delete the first chatacter of S and add it at the back of A.

Examples:

Input: S = “abab” T = “baab”
Output: YES
Explanation:
Add ‘a’ at front of A, then A = “a” and S = “bab”
Add ‘b’ at front of A, then A = “ba” and S = “ab”
Add ‘a’ at back of A, then A = “baa” and S = “b”
Add ‘b’ at back of A, then A = “baab” and S = “”
So we can make string A equal to string T

Input: S = “geeks” T = “Teeks”
Output: NO

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to use Dynamic Programming to solve this problem.

There are two possible moves for every character( front move or back move ). So, for each character, we will check if it is possible to add the character in the front or back of the new string. If it’s possible, we will move to the next character. If it’s not possible, then the operation will stop at that point and No will be printed.

1. Firstly we will make a 2D boolean array dp[][] having rows and columns equal to the length of string S, where dp[i][j] = 1 indicates that all characters of string S from index i to n-1 can be placed in the new string A with j front moves such that it becomes equal to string T.
2. We can traverse string S from the back and for each character update dp[][] in two ways, if we take the (i-1)-th character as a front move or (i-1)-th character as a back move.
3. Finally, we will check if any value at the 1st row is equal to one or not.

Below is the implementation of the above approach

 `// C++ implementation of above ` `// approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that prints whether ` `// is it possible to make a ` `// string equal to T by ` `// performing given operations ` `void` `twoStringsEquality(string s, ` `                        ``string t) ` `{ ` `    ``int` `n = s.length(); ` ` `  `    ``vector > dp( ` `        ``n, vector<``int``>( ` `               ``n + 1, 0)); ` ` `  `    ``// Base case, if we put the ` `    ``// last character at front ` `    ``// of A ` `    ``if` `(s[n - 1] == t) ` `        ``dp[n - 1] = 1; ` ` `  `    ``// Base case, if we put the ` `    ``// last character at back ` `    ``// of A ` `    ``if` `(s[n - 1] == t[n - 1]) ` `        ``dp[n - 1] = 1; ` ` `  `    ``for` `(``int` `i = n - 1; i > 0; i--) { ` `        ``for` `(``int` `j = 0; j <= n - i; j++) { ` ` `  `            ``// Condition if current ` `            ``// sequence is matchable ` `            ``if` `(dp[i][j]) { ` `                ``// Condition for front ` `                ``// move to (i - 1)th ` `                ``// character ` `                ``if` `(s[i - 1] == t[j]) ` `                    ``dp[i - 1][j + 1] = 1; ` ` `  `                ``// Condition for back ` `                ``// move to (i - 1)th ` `                ``// character ` `                ``if` `(s[i - 1] == t[i + j - 1]) ` `                    ``dp[i - 1][j] = 1; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``bool` `ans = ``false``; ` ` `  `    ``for` `(``int` `i = 0; i <= n; i++) { ` `        ``// Condition if it is ` `        ``// possible to make ` `        ``// string A equal to ` `        ``// string T ` `        ``if` `(dp[i] == 1) { ` `            ``ans = ``true``; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Print final ` `    ``// answer ` `    ``if` `(ans == ``true``) ` `        ``cout << ``"Yes"` `             ``<< ``"\n"``; ` `    ``else` `        ``cout << ``"No"` `             ``<< ``"\n"``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string S = ``"abab"``; ` ` `  `    ``string T = ``"baab"``; ` ` `  `    ``twoStringsEquality(S, T); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of above ` `// approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function that prints whether ` `// is it possible to make a ` `// String equal to T by ` `// performing given operations ` `static` `void` `twoStringsEquality(String s, ` `                               ``String t) ` `{ ` `    ``int` `n = s.length(); ` ` `  `    ``int` `[][]dp = ``new` `int``[n][n + ``1``]; ` `     `  `    ``// Base case, if we put the ` `    ``// last character at front ` `    ``// of A ` `    ``if` `(s.charAt(n - ``1``) == t.charAt(``0``)) ` `        ``dp[n - ``1``][``1``] = ``1``; ` ` `  `    ``// Base case, if we put the ` `    ``// last character at back ` `    ``// of A ` `    ``if` `(s.charAt(n - ``1``) == t.charAt(n - ``1``)) ` `        ``dp[n - ``1``][``0``] = ``1``; ` ` `  `    ``for``(``int` `i = n - ``1``; i > ``0``; i--) ` `    ``{ ` `       ``for``(``int` `j = ``0``; j <= n - i; j++) ` `       ``{ ` `            `  `          ``// Condition if current ` `          ``// sequence is matchable ` `          ``if` `(dp[i][j] > ``0``) ` `          ``{ ` `               `  `              ``// Condition for front ` `              ``// move to (i - 1)th ` `              ``// character ` `              ``if` `(s.charAt(i - ``1``) == ` `                  ``t.charAt(j)) ` `                  ``dp[i - ``1``][j + ``1``] = ``1``; ` `               `  `              ``// Condition for back ` `              ``// move to (i - 1)th ` `              ``// character ` `              ``if` `(s.charAt(i - ``1``) ==  ` `                  ``t.charAt(i + j -``1``)) ` `                  ``dp[i - ``1``][j] = ``1``; ` `          ``} ` `       ``} ` `    ``} ` ` `  `    ``boolean` `ans = ``false``; ` ` `  `    ``for``(``int` `i = ``0``; i <= n; i++) ` `    ``{ ` `         `  `       ``// Condition if it is possible ` `       ``// to make String A equal to ` `       ``// String T ` `       ``if` `(dp[``0``][i] == ``1``) ` `       ``{ ` `           ``ans = ``true``; ` `           ``break``; ` `       ``} ` `    ``} ` `     `  `    ``// Print final answer ` `    ``if` `(ans == ``true``) ` `        ``System.out.print(``"Yes"` `+ ``"\n"``); ` `    ``else` `        ``System.out.print(``"No"` `+ ``"\n"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String S = ``"abab"``; ` `    ``String T = ``"baab"``; ` ` `  `    ``twoStringsEquality(S, T); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

 `# Python3 implementation of above ` `# approach ` ` `  `# Function that prints whether ` `# is it possible to make a ` `# equal to T by ` `# performing given operations ` `def` `twoStringsEquality(s, t): ` `    ``n ``=` `len``(s) ` ` `  `    ``dp ``=` `[[``0` `for` `i ``in` `range``(n ``+` `1``)] ` `             ``for` `i ``in` `range``(n)] ` ` `  `    ``# Base case, if we put the ` `    ``# last character at front ` `    ``# of A ` `    ``if` `(s[n ``-` `1``] ``=``=` `t[``0``]): ` `        ``dp[n ``-` `1``][``1``] ``=` `1` ` `  `    ``# Base case, if we put the ` `    ``# last character at back ` `    ``# of A ` `    ``if` `(s[n ``-` `1``] ``=``=` `t[n ``-` `1``]): ` `        ``dp[n ``-` `1``][``0``] ``=` `1` ` `  `    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``): ` `        ``for` `j ``in` `range``(n ``-` `i ``+` `1``): ` ` `  `            ``# Condition if current ` `            ``# sequence is matchable ` `            ``if` `(dp[i][j]): ` `                 `  `                ``# Condition for front ` `                ``# move to (i - 1)th ` `                ``# character ` `                ``if` `(s[i ``-` `1``] ``=``=` `t[j]): ` `                    ``dp[i ``-` `1``][j ``+` `1``] ``=` `1` ` `  `                ``# Condition for back ` `                ``# move to (i - 1)th ` `                ``# character ` `                ``if` `(s[i ``-` `1``] ``=``=` `t[i ``+` `j ``-` `1``]): ` `                    ``dp[i ``-` `1``][j] ``=` `1` ` `  `    ``ans ``=` `False` ` `  `    ``for` `i ``in` `range``(n ``+` `1``): ` `         `  `        ``# Condition if it is ` `        ``# possible to make ` `        ``# A equal to T ` `        ``if` `(dp[``0``][i] ``=``=` `1``): ` `            ``ans ``=` `True` `            ``break` ` `  `    ``# Print final answer ` `    ``if` `(ans ``=``=` `True``): ` `        ``print``(``"Yes"``) ` `    ``else``: ` `        ``print``(``"No"``) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``S ``=` `"abab"` `    ``T ``=` `"baab"` ` `  `    ``twoStringsEquality(S, T) ` `     `  `# This code is contributed by mohit kumar 29     `

 `// C# implementation of above ` `// approach ` `using` `System; ` `class` `GFG{ ` ` `  `// Function that prints whether ` `// is it possible to make a ` `// String equal to T by ` `// performing given operations ` `static` `void` `twoStringsEquality(String s, ` `                               ``String t) ` `{ ` `    ``int` `n = s.Length; ` ` `  `    ``int` `[,]dp = ``new` `int``[n, n + 1]; ` `     `  `    ``// Base case, if we put the ` `    ``// last character at front ` `    ``// of A ` `    ``if` `(s[n - 1] == t) ` `        ``dp[n - 1, 1] = 1; ` ` `  `    ``// Base case, if we put the ` `    ``// last character at back ` `    ``// of A ` `    ``if` `(s[n - 1] == t[n - 1]) ` `        ``dp[n - 1, 0] = 1; ` ` `  `    ``for``(``int` `i = n - 1; i > 0; i--) ` `    ``{ ` `        ``for``(``int` `j = 0; j <= n - i; j++) ` `        ``{ ` `                 `  `            ``// Condition if current ` `            ``// sequence is matchable ` `            ``if` `(dp[i, j] > 0) ` `            ``{ ` `                     `  `                ``// Condition for front ` `                ``// move to (i - 1)th ` `                ``// character ` `                ``if` `(s[i - 1] == t[j])  ` `                    ``dp[i - 1, j + 1] = 1; ` `                     `  `                ``// Condition for back ` `                ``// move to (i - 1)th ` `                ``// character ` `                ``if` `(s[i - 1] == t[i + j - 1]) ` `                    ``dp[i - 1, j] = 1; ` `            ``} ` `        ``} ` `    ``} ` `     `  `    ``bool` `ans = ``false``; ` ` `  `    ``for``(``int` `i = 0; i <= n; i++) ` `    ``{ ` `         `  `        ``// Condition if it is possible ` `        ``// to make String A equal to ` `        ``// String T ` `        ``if` `(dp[0, i] == 1) ` `        ``{ ` `            ``ans = ``true``; ` `            ``break``; ` `        ``} ` `    ``} ` `     `  `    ``// Print readonly answer ` `    ``if` `(ans == ``true``) ` `        ``Console.Write(``"Yes"` `+ ``"\n"``); ` `    ``else` `        ``Console.Write(``"No"` `+ ``"\n"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``String S = ``"abab"``; ` `    ``String T = ``"baab"``; ` ` `  `    ``twoStringsEquality(S, T); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:
```Yes
```

Time Complexity: O(N2)

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Improved By : mohit kumar 29, 29AjayKumar

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